Sort list of dictionaries

[Peter Otten]

Charles Heizer wrote:

> On Monday, March 2, 2015 at 11:23:37 AM UTC-8, Peter Otten wrote:
>> Charles Heizer wrote:
>> 
>> > Never mind, the light bulb finally went off. :-\
>> > 
>> > sortedlist = sorted(mylist , key=lambda elem: "%s %s" % ( elem['name'],
>> > (".".join([i.zfill(5) for i in elem['version'].split(".")])) ),
>> > reverse=True)
>> 
>> This lightbulb will break with version numbers > 99999 ;)
>> 
>> Here are two alternatives:
>> 
>> result = sorted(
>>     mylist,
>>     key=lambda elem: (elem['name'], LooseVersion(elem['version'])),
>>     reverse=True)
>> 
>> result = sorted(
>>     mylist,
>>     key=lambda e: (e["name"], tuple(map(int, e["version"].split(".")))),
>>     reverse=True)
>> 
>> 
>> Personally, I prefer to not use a lambda:
>> 
>> def name_version(elem):
>>     return elem['name'], LooseVersion(elem['version'])
>>     
>> result = sorted(mylist, key=name_version, reverse=True)
> 
> Peter, thank you. Me being new to Python why don't you prefer to use a
> lambda?

I find

def name_version(elem):
    return elem['name'], LooseVersion(elem['version'])

more readable than

lambda elem: (elem['name'], LooseVersion(elem['version']))

and I can understand what

>> result = sorted(mylist, key=name_version, reverse=True)

is supposed to do without grokking the implementation of name_version() 
first. I can spot a potential bug -- are the names really supposed to occur 
in reverse order, not just the versions? -- again without a look at the 
implementation.

If I intend to use the script more than once or if I want to rely on the 
result I can write unit tests for name_version() to increase confidence that 
it does what I expect.

Finally I can add a docstring to make it more discoverable in the 
interactive interpreter.